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Zina [86]
2 years ago
5

Arrange the following compounds in order of increasing solubility in water: O2, LiCl, Br2, methanol (CH3OH).

Chemistry
1 answer:
-BARSIC- [3]2 years ago
7 0

Answer:

Methanol is more soluble > LiCl > Br2(l) > O2(g)

Explanation:

Arrange the following compounds in order to increase water solubility:

O2 = nonpolar gas, MW(Molecular weight)= 32 g/mol

LiCl= ionic salt, dissociate in Li+ and Cl-ions Br2= nonpolar liquid;

MW= 319.4 g / molmethanol(CH3OH)= polar liquid, MW= 32 g/mol

Therefore, with this data: polar methanol comes then, remember that like dissolves like, so polar + polar will dissolve pretty good LiCl appetite. However keep in mind that it will have a "saturation" point, since then most salts will not be infinitely soluble, Br2(l); since it is a non-polar liquid, it will have low solubility, but since it is a halogen, it will have low solubility.

Methanol is more soluble > LiCl > Br2(l) > O2(g)

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When an aldose reacts with Barfoed's reagent, what type of organic compound forms? What type of chemical is this?
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2 years ago
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1. A student sees tiny bubbles clinging to the inside of an unopened plastic bottle full of carbonated soft drinks. The student
Lina20 [59]

Answer:

1) The bubbles will grow, and more may appear.

2)Can A will make a louder and stronger fizz than can B.

Explanation:

When you squeeze the sides of the bottle you increase the pressure pushing on the bubble, making it compress into a smaller space. This decrease in volume causes the bubble to increase in density. When the bubble increases in density, the bubble will grow and more bubbles will appear. Therefore, Changing the pressure (by squeezing the bottle) changes the volume of the bubbles. The number of bubbles doesn't change, just their size increases.

Carbonated drinks tend to lose their fizz at higher temperatures because the loss of carbon dioxide in liquids is increased as temperature is raised. This can be explained by the fact that when carbonated liquids are exposed to high temperatures, the solubility of gases in them is decreased. Hence the solubility of CO2 gas in can A at 32°C is less than the solubility of CO2 in can B at 8°C. Thus can A will tend to make a louder fizz more than can B.

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2 years ago
What properties of metals are explained by its mobile electrons? Strength, malleability, ductility, heat conduction, current con
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Due to the presence of mobile or moving electrons in an atom they are good conductor of heat and electricity. Thus, the heat conduction and current conduction properties of metals are explained by its mobile electrons.

The other mentioned properties of metal are strength which can be explained by type of bonding within the metals, malleability explains the tendency of metals to be flattened into thin sheets, ductility explains the tendency to be stretched into wires, luster means the surface of metal is shiny and opacity is measure of impermeability that is to what extent they can pass light through them, metals are opaque, can not pass light through them or they are not transparent . All these properties are not because of mobile electrons in metals.

Therefore, correct properties are heat conduction and current conduction.


8 0
2 years ago
Read 2 more answers
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. S
Kay [80]

<u>Answer:</u> The below calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

<u>Explanation:</u>

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

We are given:

Molar mass of ^{235}UF_6=349.034348g/mol

Molar mass of ^{238}UF_6=352.041206g/mol

By taking their ratio, we get:

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{M_{(^{238}UF_6)}}{M_{(^{235}UF_6)}}}

\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\sqrt{\frac{352.041206}{349.034348}}\\\\\frac{Rate_{(^{235}UF_6)}}{Rate_{(^{238}UF_6)}}=\frac{1.00429816}{1}

From the above relation, it is clear that rate of effusion of ^{235}UF_6 is faster than ^{238}UF_6

Difference in the rate of both the gases, Rate_{(^{235}UF_6)}-Rate_{(^{238}UF_6)}=1.00429816-1=0.00429816

To calculate the percentage increase in the rate, we use the equation:

\%\text{ increase}=\frac{\Delta R}{Rate_{(^{235}UF_6)}}\times 100

Putting values in above equation, we get:

\%\text{ increase}=\frac{0.00429816}{1.00429816}\times 100\\\\\%\text{ increase}=0.4\%

The above calculations proves that the rate of diffusion of ^{235}UF_6 is 0.4 % faster than the rate of diffusion of ^{238}UF_6

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2 years ago
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